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IPS Chapter 12
© 2010 W.H. Freeman and Company
Inference for On-Way Analysis of Variance
Comparing the Means
One-Way Analysis of Variance
Objectives (IPS Chapter 12.1)
Inference for one-way ANOVA
Comparing means
The two-sample t statistic
An overview of ANOVA
The ANOVA model
Testing hypotheses in one-way ANOVA
The F-test
The ANOVA table
The idea of ANOVAReminders: A factor is a variable that can take one of several levels used to
differentiate one group from another.
An experiment has a one-way, or completely randomized, design if several
levels of one factor are being studied and the individuals are randomly
assigned to its levels. (There is only one way to group the data.)
Example: Four levels of nematode quantity in seedling growth experiment.
But two seed species and four levels of nematodes would be a two-way
design.
Analysis of variance (ANOVA) is the technique used to determine
whether more than two population means are equal.
One-way ANOVA is used for completely randomized, one-way designs.
Comparing means
We want to know if the observed differences in sample means are
likely to have occurred by chance just because of random sampling.
This will likely depend on both the difference between the sample
means and how much variability there is within each sample.
Two-sample t statistic
A two sample t-test assuming equal variance and an ANOVA comparing only
two groups will give you the exact same p-value (for a two-sided hypothesis).
H0: =
Ha: ≠
t-test assuming equal variance
t-statistic
H0: =
Ha: ≠
One-way ANOVA
F-statistic
F = t2 and both p-values are the same.
But the t-test is more flexible: You may choose a one-sided alternative instead,
or you may want to run a t-test assuming unequal variance if you are not sure
that your two populations have the same standard deviation .
We first examine the multiple populations or multiple treatments to
test for overall statistical significance as evidence of any difference
among the parameters we want to compare. ANOVA F-test
If that overall test showed statistical significance, then a detailed
follow-up analysis is legitimate.
If we planned our experiment with specific alternative hypotheses in
mind (before gathering the data), we can test them using contrasts.
If we do not have specific alternatives, we can examine all pair-wise
parameter comparisons to define which parameters differ from which,
using multiple comparisons procedures.
An Overview of ANOVA
Do nematodes affect plant growth? A botanist prepares
16 identical planting pots and adds different numbers of
nematodes into the pots. Seedling growth (in mm) is
recorded two weeks later.
Nematodes and plant growth
Nematodes0 10.8 9.1 13.5 9.2 10.65
1,000 11.1 11.1 8.2 11.3 10.435,000 5.4 4.6 7.4 5 5.6
10,000 5.8 5.3 3.2 7.5 5.45
Seedling growth
overall mean 8.03
x i
Hypotheses: All i are the same (H0)
versus not All i are the same (Ha)
The ANOVA model
Random sampling always produces chance variations. Any “factor
effect” would thus show up in our data as the factor-driven differences
plus chance variations (“error”):
Data = fit (“factor/groups”) + residual (“error”)
The one-way ANOVA model analyses
situations where chance variations are
normally distributed N(0,σ) so that:
Testing hypotheses in one-way ANOVAWe have I independent SRSs, from I populations or treatments.
The ith population has a normal distribution with unknown mean µi.
All I populations have the same standard deviation σ, unknown.
The ANOVA F statistic tests:
When H0 is true, F has the F
distribution with I − 1 (numerator)
and N − I (denominator) degrees of
freedom.
H0: = = … = I
Ha: not all the i are equal.)(SSE
)1(SSG
IN
IF
sample samein sindividual amongvariation
means sample amongvariation F
Difference in
means small
relative to overall
variability
Difference in
means large
relative to overall
variability
Larger F-values typically yield more significant results. How large depends on
the degrees of freedom (I − 1 and N − I).
The ANOVA F-statistic compares variation due to specific sources
(levels of the factor) with variation among individuals who should be
similar (individuals in the same sample).
F tends to be small F tends to be large
The ANOVA F-test
Checking our assumptions
The ANOVA F-test requires that all populations have the same
standard deviation . Since is unknown, this can be hard to check.
Practically: The results of the ANOVA F-test are approximately
correct when the largest sample standard deviation is no more than
twice as large as the smallest sample standard deviation.
(Equal sample sizes also make ANOVA more robust to deviations from the equal rule)
Each of the #I populations must be normally distributed (histograms
or normal quantile plots). But the test is robust to normality deviations
for large enough sample sizes, thanks to the central limit theorem.
Do nematodes affect plant growth?
Seedling growth si
0 nematode 10.8 9.1 13.5 9.2 10.65 2.0531000 nematodes 11.1 11.1 8.2 11.3 10.425 1.4865000 nematodes 5.4 4.6 7.4 5.0 5.6 1.24410000 nematodes 5.8 5.3 3.2 7.5 5.45 1.771
Conditions required:
• equal variances: checking that largest si no more than twice smallest si
Largest si = 2.053; smallest si = 1.244
• Independent SRSs
Four groups obviously independent
• Distributions “roughly” normal
It is hard to assess normality with only
four points per condition. But the pots in
each group are identical, and there is no
reason to suspect skewed distributions.
x i
A study of the effect of smoking classifies subjects as nonsmokers, moderate
smokers, and heavy smokers. The investigators interview a random sample of
200 people in each group and ask “How many hours do you sleep on a typical
night?”
1. Study design?
2. Hypotheses?
3. ANOVA assumptions?
4. Degrees of freedom?
1. This is an observational study.Explanatory variable: smoking -- 3 levels: nonsmokers, moderate smokers, heavy smokersResponse variable: # hours of sleep per night
2. H0: all 3 i equal (versus not all equal)
3. Three obviously independent SRS. Sample size of 200 should accommodate any departure from normality. Would still be good to check for smin/smax.
4. I = 3, n1 = n2 = n3 = 200, and N = 600, so there are I - 1 = 2 (numerator) and N - I = 597 (denominator) degrees of freedom.
Smoking influence on sleep
The ANOVA tableSource of variation Sum of squares
SSDF Mean square
MSF P value F crit
Among or between “groups”
I -1 SSG/DFG MSG/MSE Tail area above F
Value of F for
Within groups or “error”
N - I SSE/DFE
Total SST=SSG+SSE N – 1
2)( xxn ii
2)( xxij
2)1( ii sn
R2 = SSG/SST √MSE = sp
Coefficient of determination Pooled standard deviation
The sum of squares represents variation in the data: SST = SSG + SSE.
The degrees of freedom likewise reflect the ANOVA model: DFT = DFG + DFE.
Data (“Total”) = fit (“Groups”) + residual (“Error”)
Excel output for the one-way ANOVA
numeratordenominator
Here, the calculated F-value (12.08) is larger than Fcritical (3.49) for 0.05.
Thus, the test is significant at 5% Not all mean seedling lengths are
the same; the number of nematode’s is an influential factor.
ANOVA
SeedlingLength
100.647 3 33.549 12.080 .001
33.328 12 2.777
133.974 15
Between Groups
Within Groups
Total
Sum ofSquares df Mean Square F Sig.
SPSS output for the one-way ANOVA
The ANOVA found that the amount of
nematodes in pots significantly impacts
seedling growth.
The graph suggests that nematode
amounts above 1,000 per pot are
detrimental to seedling growth.
Using Table EThe F distribution is asymmetrical and has two distinct degrees of
freedom. This was discovered by Fisher, hence the label “F.”
Once again, what we do is calculate the value of F for our sample data
and then look up the corresponding area under the curve in Table E.
Fcritical for 5% is 3.49
F = 12.08 > 10.80
Thus p < 0.001
ANOVASource of Variation SS df MS F P-value F critBetween Groups 101 3 33.5 12.08 0.00062 3.4903Within Groups 33.3 12 2.78
Total 134 15
Yogurt preparation and taste
Yogurt can be made using three distinct commercial preparation
methods: traditional, ultra filtration, and reverse osmosis.
To study the effect of these methods on taste, an experiment was
designed where three batches of yogurt were prepared for each of the
three methods. A trained expert tasted each of the nine samples,
presented in random order, and judged them on a scale of 1 to 10.
Variables, hypotheses, assumptions, calculations?
ANOVA table
Source of variation SS df MS F P-value F crit
Between groups 17.3 I-1=2 8.65 11.283
Within groups 4.6 N-I=6 0.767
Total 17.769
MSG, the mean square for groups, measures how different the individual
means are from the overall mean (~ weighted average of square distances of
sample averages to the overall mean). SSG is the sum of squares for groups.
MSE, the mean square for error is the pooled sample variance sp2 and
estimates the common variance σ2 of the I populations (~ weighted average of
the variances from each of the I samples). SSE is the sum of squares for error.
)(SSE
)1(SSG
MSE
MSG
IN
IF
Computation details
MSG n1(x 1 x )2 n2(x 2 x )2 L nI (x I x )2
I 1
MSE (n1 1)s1
2 (n2 1)s22 L (nI 1)sI
2
N I
Objectives (IPS Chapter 12.2)
Comparing the means
Contrasts
Multiple comparisons
Power of the one-way ANOVA test
You have calculated a p-value for your ANOVA test. Now what?
If you found a significant result, you still need to determine which
treatments were different from which.
You can gain insight by looking back at your plots (boxplot, mean ± s).
There are several tests of statistical significance designed specifically for
multiple tests. You can choose apriori contrasts, or aposteriori multiple
comparisons.
You can find the confidence interval for each mean i shown to be
significantly different from the others.
Contrasts can be used only when there are clear expectations BEFORE starting an experiment, and these are reflected in the experimental design. Contrasts are planned comparisons.
Patients are given either drug A, drug B, or a placebo. The three treatments are not symmetrical. The placebo is meant to provide a baseline against which the other drugs can be compared.
Multiple comparisons should be used when there are no justified expectations. Those are aposteriori, pair-wise tests of significance.
We compare gas mileage for eight brands of SUVs. We have no prior knowledge to expect any brand to perform differently from the rest. Pair-wise comparisons should be performed here, but only if an ANOVA test on all eight brands reached statistical significance first.
It is NOT appropriate to use a contrast test when suggested comparisons appear only after the data is collected.
Contrasts: planned comparisons When an experiment is designed to test a specific hypothesis that
some treatments are different from other treatments, we can use
contrasts to test for significant differences between these specific
treatments.
Contrasts are more powerful than multiple comparisons because they are
more specific. They are more able to pick up a significant difference.
You can use a t-test on the contrasts or calculate a t-confidence interval.
The results are valid regardless of the results of your multiple sample
ANOVA test (you are still testing a valid hypothesis).
A contrast is a combination of
population means of the form :
Where the coefficients ai have sum 0.
The corresponding sample contrast is :
The standard error of c is :
iia
ii xac
cSEct
i
i
i
ipc n
aMSE
n
asSE
22
To test the null hypothesis H0: = 0 use the t-statistic:
With degrees of freedom DFE that is associated with sp. The alternative hypothesis can be one- or two-sided.
cSEtc *
A level C confidence interval for the difference is :
Where t* is the critical value defining the middle C% of the t distribution with DFE degrees of freedom.
Contrasts are not always readily available in statistical software
packages (when they are, you need to assign the coefficients “ai”), or
may be limited to comparing each sample to a control.
If your software doesn’t provide an option for contrasts, you can test
your contrast hypothesis with a regular t-test using the formulas we just
highlighted. Remember to use the pooled variance and degrees of
freedom as they reflect your better estimate of the population variance.
Then you can look up your p-value in a table of t-distribution.
Do nematodes affect plant growth? A botanist prepares
16 identical planting pots and adds different numbers of
nematodes into the pots. Seedling growth
(in mm) is recorded two weeks later.
Nematodes and plant growth
Nematodes0 10.8 9.1 13.5 9.2 10.65
1,000 11.1 11.1 8.2 11.3 10.435,000 5.4 4.6 7.4 5 5.6
10,000 5.8 5.3 3.2 7.5 5.45
Seedling growth
overall mean 8.03
x i
One group contains no nematode at all. If the botanist planned this group as a
baseline/control, then a contrast of all the nematode groups against the
control would be valid.
Nematodes: planned comparison
475.1045.56.5425.1065.10*3 ii xac
Contrast of all the nematode groups against the control:
Combined contrast hypotheses:
H0: µ1 = 1/3 (µ2+ µ3 + µ4) vs.
Ha: µ1 > 1/3 (µ2+ µ3 + µ4) one tailed
Contrast coefficients: (+1 −1/3 −1/3 −1/3) or (+3 −1 −1 −1)
9.24
)1(*3
4
3*78.2
222
i
ipc n
asSE
12 :df 6.39.25.10 N-ISEct c
In Excel: TDIST(3.6,12,1) = tdist(t, df, tails) ≈ 0.002 (p-value).
Nematodes result in significantly shorter seedlings (alpha 1%).
x¯i si
G1: 0 nematode 10.65 2.053G2: 1,000 nematodes 10.425 1.486G3: 5,000 nematodes 5.6 1.244G4: 1,0000 nematodes 5.45 1.771
ANOVA
SeedlingLength
100.647 3 33.549 12.080 .001
33.328 12 2.777
133.974 15
Between Groups
Within Groups
Total
Sum ofSquares df Mean Square F Sig.
Contrast Coefficients
-3 1 1 1Contrast1
0 1000 5000 10000
NematodesLevel
Contrast Tests
-10.4750 2.88650 -3.629 12 .003
-10.4750 3.34823 -3.129 4.139 .034
Contrast1
1
Assume equal variances
Does not assume equalvariances
SeedlingLength
Value ofContrast Std. Error t df Sig. (2-tailed)
Planned comparison:
H0: µ1 = 1/3 (µ2+ µ3 + µ4) vs.
Ha: µ1 > 1/3 (µ2+ µ3 + µ4) one tailed
Contrast coefficients: (+3 −1 −1 −1)
ANOVA: H0: all µi are equal vs. Ha: not all µi are equal
ANOVA vs. contrasts in SPSS
not all µi are equal
Nematodes result in significantly shorter seedlings (alpha 1%).
Multiple comparisonsMultiple comparison tests are variants on the two-sample t-test.
They use the pooled standard deviation sp = √MSE.
The pooled degrees of freedom DFE.
And they compensate for the multiple comparisons.
We compute the t-statistic for all pairs of means:
A given test is significant (µi and µj significantly different), when
|tij| ≥ t** (df = DFE).
The value of t** depends on which procedure you choose to use.
tij (x i x j )
sp
1ni
1n j
The Bonferroni procedure performs a number of pair-wise
comparisons with t-tests and then multiplies each p-value by the
number of comparisons made. This ensures that the probability of
making any false rejection among all comparisons made is no greater
than the chosen significance level α.
As a consequence, the higher the number of pair-wise comparisons you
make, the more difficult it will be to show statistical significance for each
test. But the chance of committing a type I error also increases with the
number of tests made. The Bonferroni procedure lowers the working
significance level of each test to compensate for the increased chance of
type I errors among all tests performed.
The Bonferroni procedure
Simultaneous confidence intervals
We can also calculate simultaneous level C confidence intervals for
all pair-wise differences (µi − µj) between population means:
jipji nn
stxxCI11
**)(:
sp is the pooled variance, MSE.
t** is the t critical with degrees of freedom DFE = N – I, adjusted for
multiple, simultaneous comparisons (e.g., Bonferroni procedure).
SYSTAT File contains variables: GROWTH NEMATODES$Categorical values encountered during processing are:NEMATODES$ (four levels): 10K, 1K, 5K, none
Dep Var: GROWTH N: 16 Multiple R: 0.867 Squared multiple R: 0.751 Analysis of VarianceSource Sum-of-Squares df Mean-Square F-ratio PNEMATODES$ 100.647 3 33.549 12.080 0.001Error 33.328 12 2.777
Post Hoc test of GROWTH Using model MSE of 2.777 with 12 df. Matrix of pairwise mean differences:
1 2 3 41 0.0002 4.975 0.0003 0.150 −4.825 0.0004 5.200 0.225 5.050 0.000
Bonferroni AdjustmentMatrix of pairwise comparison probabilities:
1 2 3 41 1.0002 0.007 1.0003 1.000 0.009 1.0004 0.005 1.000 0.006 1.000
SigmaStat—One-Way Analysis of Variance
Normality Test: Passed (P > 0.050)Equal Variance Test: Passed (P = 0.807)
Group Name N Missing Mean Std dev SEMNone 4 0 10.650 2.053 1.0271K 4 0 10.425 1.486 0.7435K 4 0 5.600 1.244 0.62210K 4 0 5.450 1.771 0.886
Source of variation DF SS MS F P Between groups 3 100.647 33.549 12.080 <0.001Residual 12 33.328 2.777Total 15 133.974
Power of performed test with alpha = 0.050: 0.992
All Pairwise Multiple Comparison Procedures (Bonferroni t-test):Comparisons for factor: NematodesComparison Diff of means t P P<0.050None vs. 10K 5.200 4.413 0.005 YesNone vs. 5K 5.050 4.285 0.006 YesNone vs. 1K 0.225 0.191 1.000 No1K vs. 10K 4.975 4.222 0.007 Yes1K vs. 5K 4.825 4.095 0.009 Yes5K vs. 10K 0.150 0.127 1.000 No
Multiple Comparisons
Dependent Variable: SeedlingLength
Bonferroni
.22500 1.17841 1.000 -3.4901 3.9401
5.05000* 1.17841 .006 1.3349 8.7651
5.20000* 1.17841 .005 1.4849 8.9151
-.22500 1.17841 1.000 -3.9401 3.4901
4.82500* 1.17841 .009 1.1099 8.5401
4.97500* 1.17841 .007 1.2599 8.6901
-5.05000* 1.17841 .006 -8.7651 -1.3349
-4.82500* 1.17841 .009 -8.5401 -1.1099
.15000 1.17841 1.000 -3.5651 3.8651
-5.20000* 1.17841 .005 -8.9151 -1.4849
-4.97500* 1.17841 .007 -8.6901 -1.2599
-.15000 1.17841 1.000 -3.8651 3.5651
(J) NematodesLevel1000
5000
10000
0
5000
10000
0
1000
10000
0
1000
5000
(I) NematodesLevel0
1000
5000
10000
MeanDifference
(I-J) Std. Error Sig. Lower Bound Upper Bound
95% Confidence Interval
The mean difference is significant at the .05 level.*.
SPSSANOVA
SeedlingLength
100.647 3 33.549 12.080 .001
33.328 12 2.777
133.974 15
Between Groups
Within Groups
Total
Sum ofSquares df Mean Square F Sig.
The ANOVA test is significant (α 5%): we have found evidence that the three
methods do not all yield the same population mean reading comprehension score.
Teaching methods
A study compares the reading
comprehension (“COMP,” a test
score) of children randomly
assigned to one of three teaching
methods: basal, DRTA, and
strategies.
We test: H0: µBasal = µDRTA = µStrat vs. Ha: H0 not true
What do you conclude?
The three methods do not yield the same results: We found evidence of a
significant difference between DRTA and basal methods (DRTA gave better
results on average), but the data gathered does not support the claim of a
difference between the other methods (DRTA vs. strategies or basal vs.
strategies).
PowerThe power, or sensitivity, of a one-way ANOVA is the probability that
the test will be able to detect a difference among the groups (i.e. reach
statistical significance) when there really is a difference.
Estimate the power of your test while designing your experiment to
select sample sizes appropriate to detect an amount of difference
between means that you deem important.
Too small a sample is a waste of experiment, but too large a sample is
also a waste of resources.
A power of at least 80% is often suggested.
Power computations
ANOVA power is affected by
The significance level
The sample sizes and number of groups being compared
The differences between group means µi
The guessed population standard deviation
You need to decide what alternative Ha you would consider important,
detect statistically for the means µi, and to guess the common standard
deviation σ (from similar studies or preliminary work).
The power computations then require calculating a non-centrality
paramenter λ, which follows the F distribution with DFG and DFE
degrees of freedom to arrive at the power of the test.
Systat: Power analysis Alpha = 0.05
Model = Oneway
Number of groups = 4
Within cell S.D. = 1.5
Mean(01) = 7.0
Mean(02) = 8.0
Mean(03) = 9.0
Mean(04) = 10.0
Effect Size = 0.745
Noncentrality parameter = 2.222 * sample size
SAMPLE SIZE POWER
(per cell)
3 0.373
4 0.551
5 0.695
6 0.802
7 0.876
8 0.925
If we anticipated a gradual decrease of
seedling length for increasing amounts
of nematodes in the pots and would
consider gradual changes of 1 mm on
average to be important enough to be
reported in a scientific journal…
…then we would reach a power of 80%
or more when using six pots or more for
each condition.
(Four pots per condition would only
bring a power of 55%.)
guessed
Systat: Power analysis Alpha = 0.05
Power = 0.80
Model = One-way
Number of groups = 4
Within cell S.D. = 1.7
Mean(01) = 6.0
Mean(02) = 6.0
Mean(03) = 10.0
Mean(04) = 10.0
Effect size = 1.176
Noncentrality parameter = 5.536 * sample size
SAMPLE SIZE POWER (per cell)
2 0.394
3 0.766
4 0.931
Total sample size = 16
If we anticipated that only large amounts
of nematodes in the pots would lead to
substantially shorter seedling lengths
and would consider step changes of 4
mm on average an important effect…
…then we would reach a power of 80%
or more when using four pots or more in
each group (three pots per condition
might be close enough though).
guessed